Use the Question 11 Writing Booklet (a) Let $P(x) = x^3 + 3x^2 - 13x + 6$. (i) Show that $P(2) = 0$. (ii) Hence, factor the polynomial $P(x)$ as $A(x)B(x)$, where $B(x)$ is a quadratic polynomial. (b) For what value(s) of $a$ are the vectors $\begin{pmatrix} a \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 2a - 3 \\ 2 \end{pmatrix}$ perpendicular? (c) The diagram shows the graph of $y = f(x)$. Sketch the graph of $y = \frac{1}{f(x)}$.

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Use the Question 11 Writing Booklet (a) Let $P(x) = x^3 + 3x^2 - 13x + 6$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2020 - Paper 1

Question 11

Use the Question 11 Writing Booklet (a) Let $P(x) = x^3 + 3x^2 - 13x + 6$. (i) Show that $P(2) = 0$. (ii) Hence, factor the polynomial $P(x)$ as $A(x)B(x)$, where... show full transcript

Worked Solution & Example Answer:Use the Question 11 Writing Booklet (a) Let $P(x) = x^3 + 3x^2 - 13x + 6$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2020 - Paper 1

Step 1

Show that $P(2) = 0$.

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Answer

To evaluate $P(2)$ , substitute $x = 2$ into the polynomial:

P(2) = 2^3 + 3(2^2) - 13(2) + 6 = 8 + 12 - 26 + 6 = 0.

Thus, $P(2) = 0$ . This confirms that 2 is a root of the polynomial.

Step 2

Hence, factor the polynomial $P(x)$ as $A(x)B(x)$, where $B(x)$ is a quadratic polynomial.

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Answer

Given that $P(2) = 0$ , we can factor $P(x)$ as follows:

Using synthetic division to divide $P(x)$ by $(x - 2)$ :

egin{array}{r|rrrr} 2 & 1 & 3 & -13 & 6 \\ & & 2 & 10 & -6 \\ ext{------------------} \\ & 1 & 5 & -3 & 0 \\ ext{Thus, } \ P(x) = (x - 2)(x^2 + 5x - 3).

So, we have factored the polynomial as $A(x) = (x - 2)$ and $B(x) = (x^2 + 5x - 3)$ .

Step 3

For what value(s) of $a$ are the vectors $\begin{pmatrix} a \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 2a - 3 \\ 2 \end{pmatrix}$ perpendicular?

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Answer

Vectors are perpendicular if their dot product is zero:

\begin{pmatrix} a \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2a - 3 \\ 2 \end{pmatrix} = 0.

Calculating the dot product:

a(2a - 3) + (-1)(2) = 0 \\ 2a^2 - 3a - 2 = 0.

Next, we can solve this quadratic using the quadratic formula:

a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)}.

This simplifies to:

a = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}.

Thus, the solutions are:

a = 2 \text{ or } -\frac{1}{2}.

Step 4

Sketch the graph of $y = \frac{1}{f(x)}$.

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Answer

To sketch the graph of $y = \frac{1}{f(x)}$ , we first need to understand the behavior of $f(x)$ , which is a downward-opening parabola with a minimum point.

At the minima, the value of $f(x)$ approaches zero, leading to vertical asymptotes in $y = \frac{1}{f(x)}$ . The asymptotes occur at the $x$ -values where $f(x) = 0$ , which we found earlier occurs at $x = 2$ .

Also, as $x$ approaches the minima of $f(x)$ , $y = \frac{1}{f(x)}$ approaches infinity. Between the roots and the location of the minima, the graph of $y = \frac{1}{f(x)}$ will exhibit a downward curve.

Therefore, the sketch will have the necessary features including vertical asymptotes as described.

Use the Question 11 Writing Booklet (a) Let $P(x) = x^3 + 3x^2 - 13x + 6$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2020 - Paper 1

Worked Solution & Example Answer:Use the Question 11 Writing Booklet (a) Let $P(x) = x^3 + 3x^2 - 13x + 6$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2020 - Paper 1

Show that $P(2) = 0$.

Hence, factor the polynomial $P(x)$ as $A(x)B(x)$, where $B(x)$ is a quadratic polynomial.

For what value(s) of $a$ are the vectors $\begin{pmatrix} a \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 2a - 3 \\ 2 \end{pmatrix}$ perpendicular?

Sketch the graph of $y = \frac{1}{f(x)}$.

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